# Let $S=\{v_1,v_2\}$ be linearly independent. If $[S] \ne V$, then there is a vector $v_3 \in V$ such that $\{v_1,v_2,v_3\}$ is linearly independent.

I couldn't fit the whole question in the title, so here is what I was asked: Suppose $$S = \{ v_1, v_2\}$$ is a linearly independent set in a finite dimensional vector space $$V$$. Show that if $$[S] \ne V$$, then there is a vector $$v_3 \in V$$ such that $$\{v_1,v_2,v_3\}$$ is linearly independent.

I tried visualizing this in the $$\mathbb{R}^3$$ vector space, and it makes sense because the span of $$S$$ would be the vectors that are on a plane in the 3-dimensional space, and if you introduce a new vector so that the three vectors are now linearly independent, then the vector must be outside of that plane. But I'm struggling to formalize this into words.

I tried proving this statement by the contrapositive: If $$\forall v_3\in V$$ $$\{v_1,v_2,v_3\}$$ is linearly dependent, then $$[S]=V$$. This is how I did it so far:

Suppose that for all vectors $$v_3\in V$$, the set $$\{v_1,v_2,v_3\}$$ is linearly dependent. This means that there is a nontrivial solution $$\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$$ to the equation

\begin{align*} \alpha_1v_1+\alpha_2v_2+\alpha_3v_3=\Theta \end{align*}

However, we know that $$S$$ is linearly independent, so \begin{align*} c_1v_1+c_2v_2=\Theta\end{align*} where $$c_1=c_2=0$$. Therefore, we have that $$\alpha_1=\alpha_2=0$$. But since there must be a nontrivial solution to the equation, we know $$|\alpha_1|+|\alpha_2|+|\alpha_3|\neq0\neq0+0+|\alpha_3|$$. Therefore, \begin{align*} 0v_1+0v_2+a_3v_3=\Theta\\ \Theta+\Theta+a_3v_3=\Theta\\ a_3v_3=\Theta \end{align*} Since $$a_3\neq0$$, $$v_3$$ must be $$\Theta$$.

Once I got up to this point, I started feeling like I made an incorrect assumption somewhere or that I'm going in the wrong direction. Is it true, given my initial assumption, that $$v_3=\Theta$$? If not, how can I show that $$[S]=V$$? Or maybe my entire approach is incorrect. I'd appreciate any pointers.

Using the contrapositive approach: if $$v_3\in V$$ is arbitrary and $$v_1,v_2,v_3$$ are linearly dependent, then as you note there are scalars $$\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$$ not all zero with $$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3=0$$ If $$\alpha_3=0$$, then this would imply that $$\alpha_1v_1+\alpha_2v_2=0$$ where $$\alpha_1,\alpha_2$$ are not both zero, contradicting the assumption that $$v_1,v_2$$ are linearly independent. So we must have $$\alpha_3\ne0$$, and therefore $$v_3=-\frac{\alpha_1}{\alpha_3}v_1-\frac{\alpha_2}{\alpha_3}v_2$$ which shows that $$v_3$$ is in the span of $$v_1,v_2$$. Since $$v_3$$ was arbitrary, this means $$v_1,v_2$$ span $$V$$.